One of the fascinating and practically important topics in physics is the thermodynamics of gases. Systems in this state of aggregation are considered using the so-called ideal gas model. This article is devoted to isoprocesses in physics studied in grade 10 and solving problems on this topic.
Ideal gas model and equation of state
As you know, gases are high-energy systems that can easily change their shape and volume. The ideal gas model is that it assumes the lack of particle size and the interaction between them. The validity of this model is obvious in a number of practical cases, since the distances between gas molecules are much larger than their linear dimensions, and the kinetic energy of their motion is several orders of magnitude greater than the weak interactions between chemically neutral particles.
In the 30s of the XIX century, collecting and analyzing all available experimental material, the French scientist Clapeyron wrote down a universal equation describing the state of any ideal gas system. This equation has the form:
P * V = n * R * T.
It can be seen that the equality contains all three basic thermodynamic variables (P - pressure, V - volume and T - temperature). In addition, it does not depend on the chemical properties of the system, but is determined only by the amount of substance n in it. The symbol R denotes the gas constant equal to 8.314 J / (mol * K).
Isothermal process
We begin to study isoprocesses in physics by considering the transition of a system between two equilibrium states at a constant temperature. If the system is closed, that is, it can exchange heat with the environment, but not with the substance, then the right-hand side of the equality in the above equation of state will be a constant value. This fact leads to the following isothermal law:
P * V = const at T = const.
This equality is called the Boyle-Marriott law in honor of the names of scientists who first discovered it experimentally in the 17th century. If we represent the isotherm in the axes P and V, then it will be a hyperbola, shown below in the figure.
Example problem for the Boyle-Mariotte law
We solve the following simple problem in physics for isoprocesses (Grade 10):
It is known that the gas under the piston expands isothermally. What will be the final pressure in the system if at the beginning it was one atmosphere? The volume as a result of gas expansion increased by 3 times.
We write the Boyle-Marriott law in this form:
P 1 * V 1 = P 2 * V 2 .
We transform this equality so that the same physical quantities appear in one part of the equality:
P 2 / P 1 = V 1 / V 2 .
Where do we get the final expression for P 2 and the answer to the question of the problem:
P 2 = P 1 * V 1 / V 2 = 1 * 1/3 = 0.33 atm.
The final pressure in the system is 3 times less than the initial.
Isobaric and isochoric processes
Both processes were specifically put forward in one paragraph of the article, since they are described by mathematical expressions of the same form.
If the gas system expands or contracts, while maintaining a constant pressure, then we speak of an isobaric process. In this case, use the following formula:
V / T = const at P = const.
This expression can be obtained from the equation of state. It is called Charlesโs law (the name of the Frenchman who discovered the law at the end of the 18th century).
Now suppose we started heating gas in a solid, closed vessel. The volume will remain constant. What will happen to temperature and pressure in the system? The experiments and the equation of state say that these quantities will grow in proportion to each other, that is:
P / T = const at V = const.
The described process is called isochoric, and the law reflecting it is named after the Frenchman Gay-Lussac, who was the first to study it experimentally and publish it at the beginning of the 19th century.
From the presented formulas it follows that their graphic representation in the axes VT (isobar) and PT (isochore) is a straightforward relationship. An example of an isobar is shown below.
Charles problem law example
Consider the following physics problem for isoprocesses:
In one cylinder, under the movable piston, 2 mol of some gas is contained at a pressure of 1 atmosphere. In another cylinder there is gas in the amount of substance 3 mol, the pressure in it is 1.5 atmospheres. Both cylinders began to heat at constant pressure. How will the isobar lines on the graph in the VT axes be mutually arranged for the processes in each cylinder?
To answer the question posed, you should write Charles's law in this form:
V = const * T.
This means that the slope for each line is equal to the value of the constant. If it is found for processes in each cylinder, then we will get the answer to the problem. To do this, we use the equation of state, we have:
P * V = n * R * T =>
V / T = n * R / P.
The right side of the last equality is the constant that we are looking for. As you can see, it depends in direct proportion to the amount of matter in the system and inversely to pressure. Substituting these values โโfrom the condition, we obtain the values โโof the constants for each process:
const1 = n 1 * R / P 1 = 2/1 * R = 2 * R;
const2 = n 2 * R / P 2 = 3 / 1,5 * R = 2 * R.
Thus, the values โโof the constants for both processes are equal, which means complete coincidence of the isobar graphs. Note that when calculating the constants, we substituted the pressure in the atmospheres. The equality of const1 and const2 will not change if we substitute the pressure in pascals.
Example Gay-Lussac Law Problem
As shown, the successful solution of problems in physics for isoprocesses requires knowledge not only of the processes themselves, but also of the ability to use the equation of state for them. We give an example of another similar problem.
Oxygen weighing 20 grams is in a closed vessel of 2 liters. Cooling the vessel by 10 degrees Celsius led to a drop in pressure in the vessel by a certain amount. It is necessary to determine the final equilibrium pressure in the system, if at the beginning it was equal to one atmosphere.
The problem is about the isochoric transition of gaseous oxygen. We get a working formula for him. To do this, we express the ratio of pressure to temperature from the equation of state:
P / T = n * R / V = โโm * R / (M * V).
Where the amount of substance n is written as the ratio of the mass of gas m to its molar mass M. Using this expression, we determine the difference between the final and initial pressures, we obtain:
P 2 - P 1 = m * R / (M * V) * (T 2 - T 1 ).
Where do we come to the final expression that allows us to answer the question posed in the problem:
P 2 = P 1 + m * R / (M * V) * ฮT, where ฮT = T 2 - T 1 .
Substitute all known quantities in the formula in SI units. After substituting and performing the calculations, we come to the answer: P 2 = 75343.75 Pa, which is equal to a pressure of 0.74 atmosphere.